I've been stuck on this for ages, and can't figure this out.

What is the missing box, and the logic behind the answer?

enter image description here

This was taken from a Korn Ferry Leadership Assessment practice trial.

  • $\begingroup$ I was able to come up with the correct answer, and with a much, much simpler reason than the other answers. I'd love to share (and get feedback), it's not worth its own answer and I don't think there's a way to hide spoilers in comments... :( $\endgroup$ – Steve Apr 14 '19 at 21:43
  • $\begingroup$ I'll post it as an answer anyway, I can always delete it. $\endgroup$ – Steve Apr 14 '19 at 21:43
  • $\begingroup$ Rrz0, please edit in where this puzzle came from. See the close notice for more info. $\endgroup$ – Rubio Apr 15 '19 at 5:49
  • $\begingroup$ @Rubio, apologies for this, will update. Do let me know if further action should be taken from my end. $\endgroup$ – rrz0 Apr 15 '19 at 6:55
  • $\begingroup$ Thanks Rrz0. Reopened. $\endgroup$ – Rubio Apr 15 '19 at 7:47

I think the answer is



If you look at each diagonal going from top left to bottom right, there appears to be a pattern of opposing ratios in colour, and corresponding ratios in edges.

Let $\rm B : W$ be the ratio of Black shapes to White shapes, then the first diagonal (bottom left corner) is $1 : 3$.

Next diagonal (left middle box and bottom middle box) is $4 : 0$ and $0 : 4$ respectively.

Next diagonal (the middle diagonal with the missing box) have the ratios $1 : 3$ and $3 : 1$, so to me it makes sense that the missing box has a ratio of $1 : 3$.

That leaves either

A or F

Due to how the shapes in these boxes are positioned, I am leaning towards


But also, in order to explain the colour scheme...

...it has to do with the edges! Here is how we will count them for each shape:

White circle's edges: 0
Black circle's edges: 1
Donut's edges: 2 (because the outer edge and inner edge are accounted for since the donut is coloured black, because the circle coloured black has its outer edge accounted for as well).
All other shapes have the number of edges you see they have.

And then one last rule:

Every shape corresponds to another shape with 4+ the edges (with alternating colour).

Let me explain:

The top left box has a black circle in the top right and the rest are white circles. That leaves a $1 : 0$ edge ratio (in the same ratio form as $\rm B : W$ but not in terms of colour anymore). Therefore, the next box in this diagonal (the middle box) must share this same edge ratio.

All the white circles correspond to black diamonds (white circles have 0 edges, 0 + 4 = 4 so the next shape has 4 edges, and the opposite of white is black, so we have black diamonds). So, in the top left box, we replace the white circles with black diamonds, and then change the black circle to a white pentagon (black circles have 1 edge, 1 + 4 = 5 so the next shape has 5 edges, and the opposite of black is white, so this makes the next shape a white pentagon). Now we can maintain our edge ratio, and violá! Lo and behold the box in the middle!

Top left $\LARGE\stackrel{\implies}{\vphantom{\prod_{n=1}^\Uparrow}}$ Middle
white circles (0 edges) $\to$ black diamonds (0 + 4 edges);
black circles (1 edge) $\to$ white pentagons (1 + 4 edges).

Same concept applies for the first parallel diagonals above and below the middle diagonal, if you want to try this pattern with the other boxes!

Get it, now?

Repeating this process in the middle diagonal from our middle box, black diamonds go to white octagons (black 4 $\to$ white $8$) and white pentagons go to black nonagons (white 5 $\to$ black $9$).

Middle $\LARGE\stackrel{\implies}{\vphantom{\prod_{n=1}^\Uparrow}}$ Answer A
Recognise the pattern? :)

This is true

for every diagonal. You can count the bottom left corner and top right corner as part of the same diagonal, and the pattern still works.

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    $\begingroup$ Yay! Thanks @Rrz0 for the $\color{green}{\checkmark}$! It was a nice puzzle :D $\endgroup$ – Mr Pie Apr 14 '19 at 17:33
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    $\begingroup$ Thank you @user477343 for the time you took to solve the puzzle. $\endgroup$ – rrz0 Apr 14 '19 at 21:03
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    $\begingroup$ Please remember, we require attribution for puzzles that are not the creation of the person posting them here. There is a close reason for that; it's there for a reason; please use it. If you don't have close privilege, you can flag it for the same reason. We don't want to encourage people posting unattributed content by providing answers, so .... close or flag, don't answer. Thanks! $\endgroup$ – Rubio Apr 15 '19 at 5:47
  • $\begingroup$ @Rubio do you wish for me to delete my answer? I can do that. I flagged the question, albeit it has already been closed. $\endgroup$ – Mr Pie Apr 15 '19 at 5:53
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    $\begingroup$ Nah, you can leave it; just a note for next time. Thanks :) $\endgroup$ – Rubio Apr 15 '19 at 5:58

I don't think this is really an answer. By only looking at the question I arrived at the right answer. It could be luck, but I think the puzzle might be improved slightly...

There is a simple pattern, if the boxes are "read" left-to-right, top-to-bottom. The first one contains shapes containing maximum one side. Block 3 has triangles, 4 has squares, etc. I lazily justified that the second box was "weird" because lines aren't used, but circles in circles ... two sides?

The choices to select an answer from only contain a single selection with a 9-sided polygon.

I'd recommend adding a nonagon to at least one other of the choices :)

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    $\begingroup$ Thanks for your contribution. I think this directly relates to my rule number 2. $\endgroup$ – rrz0 Apr 15 '19 at 6:58

Thank you @user477343 for solving the puzzle.

I found another(?) pattern. In general the puzzle above follows 2 simple rules:

Rule 1:

Each box increases in the number of shaded shapes in a clock-wise fashion. When all four shapes are shaded, each following box increases in the number of un-shaded shapes in a clock-wise manner.

Let us take the shaded circle in box 1 is the starting point.

(Box 1 to 4) show an increase in shaded number of shapes and then from Boxes 5 to 8 there is a decrease in the number of shaded shapes,in a clock-wise manner. Therefore, by simply following this rule, the answer would be A since it has one shaded shape, which is also in the correct starting position.

Rule 2:

With each increase in shaded shape, there is an also an increase in the number of sides (+1) that the newly shaded shapes have.

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    $\begingroup$ I think my answer is a consequence of your answer; i.e. yours is actually the intended one. Ha! I suppose I was merely overthinking. $(+1)$ $\endgroup$ – Mr Pie Apr 15 '19 at 3:50
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    $\begingroup$ @user477343 .... please stop the mathjax +1s, and +1s in general. Your vote suffices. $\endgroup$ – Rubio Apr 15 '19 at 7:50
  • $\begingroup$ @Rubio oh s***! Sorry, sorry. I am just so used to it, especially after writing such comments on the Math Stack Exchange. No more!!! $\endgroup$ – Mr Pie Apr 15 '19 at 8:01

The other answers have given several reasons for why the answer is A, rather than another one of the choices. But we can do better and show that it has to be the drawing in A, and can't be anything else (that's not even in the choices):

The pattern is in reading direction (left to right, then top to bottom). At each step, we toggle the color of one of the four shapes (black <-> white), going clockwise (bottom-right, bottom-left, top-left, top-right, ...). Then, we take all the shapes of the same color as the one we just toggled, and add a side to each.

So to get the missing cell, we toggle the top-right shape, and since it becomes the only black one, we add a side to only that one, giving us A.

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